(|x+3|)+4*x>=6 inequation

Given the inequality:
$$4 x + \left|{x + 3}\right| \geq 6$$
To solve this inequality, we must first solve the corresponding equation:
$$4 x + \left|{x + 3}\right| = 6$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$x + 3 \geq 0$$
or
$$-3 \leq x \wedge x < \infty$$
we get the equation
$$4 x + \left(x + 3\right) - 6 = 0$$
after simplifying we get
$$5 x - 3 = 0$$
the solution in this interval:
$$x_{1} = \frac{3}{5}$$

2.
$$x + 3 < 0$$
or
$$-\infty < x \wedge x < -3$$
we get the equation
$$4 x - \left(x + 3\right) - 6 = 0$$
after simplifying we get
$$3 x - 9 = 0$$
the solution in this interval:
$$x_{2} = 3$$
but x2 not in the inequality interval


$$x_{1} = \frac{3}{5}$$
$$x_{1} = \frac{3}{5}$$
This roots
$$x_{1} = \frac{3}{5}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{3}{5}$$
=
$$\frac{1}{2}$$
substitute to the expression
$$4 x + \left|{x + 3}\right| \geq 6$$
$$4 \cdot \frac{1}{2} + \left|{\frac{1}{2} + 3}\right| \geq 6$$

11/2 >= 6

but

11/2 < 6

Then
$$x \leq \frac{3}{5}$$
no execute
the solution of our inequality is:
$$x \geq \frac{3}{5}$$

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